For angles $A$ and $B$ in a triangle, is $\cos\frac B2-\cos \frac A2=\cos B-\cos A$ enough to conclude that $A=B$?

Question

Brief enquiry:

$$\cos\frac B2-\cos \frac A2=\cos B-\cos A$$

Optionally

$$\sqrt\frac{1+\cos B}{2}-\cos B=\sqrt\frac{1+\cos A}{2}-\cos A$$

Is above equality sufficient to prove that it implies $A=B$?

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Detailed explanation and motivation for this question:

Consider a triangle with bisectors of equal length:

By definition:

$\lvert AE\rvert = \lvert BD\rvert = D\\\frac A2+\frac A2 = A,\space\space\space\frac B2 + \frac B2 = B$

By cosine law: $$\lvert AE\rvert^2=y^2+a^2-2ya\cos B\\\lvert BD\rvert^2=x^2+a^2-2xa\cos A\\x^2=\lvert BD\rvert^2+a^2-2\lvert BD\rvert a\cos \frac B2\\y^2=\lvert AE\rvert^2+a^2-2\lvert AE\rvert a\cos \frac A2$$ By sine law: $$\frac{x}{D}=\frac{\sin \frac B2}{\sin A};\space\space\space \frac{y}{D}=\frac{\sin \frac A2}{\sin B}$$ $$\bigl[D=\lvert AE\rvert = \lvert BD\rvert\bigr]$$ Since bisectors are equal: $$x^2-2xa\cos A=y^2-2ya\cos B\implies \lvert BD\rvert \cos \frac B2 +x\cos A = \lvert AE\rvert \cos \frac A2 +y\cos B $$

$$D (\cos\frac B2-\cos \frac A2)=y\cos B-x\cos A $$

Dividing by D and substituting y and x we obtain: $$\cos\frac B2-\cos \frac A2=\frac{\sin \frac A2\cos B}{\sin B}-\frac{\sin \frac B2\cos A}{\sin A}$$

Consider triangles $\Delta$ABE and $\Delta$BAD

Area of triangle $\Delta$ABE: $$A = \frac{aD}{2}\sin\frac A2 = \frac{aD}{2}\sin B \implies \sin\frac A2 = \sin B$$

Similarly for triangle $\Delta$BAD $$A=\frac{aD}{2}\sin\frac B2 = \frac{aD}{2}\sin A\implies \sin\frac B2 = \sin A$$

Therefore:

$$\cos\frac B2-\cos \frac A2=\cos B-\cos A$$

Optionally

$$\sqrt\frac{1+\cos B}{2}-\cos B=\sqrt\frac{1+\cos A}{2}-\cos A$$

Is above equality sufficient to prove that it implies A = B?

Answer 1

After more thinking, I realized that the answer is in fact "Yes".

As before, let $a=\cos\frac{A}{2},b=\cos\frac{B}{2}, 0<a,b<1$

The identity becomes:

$2a^2-a=2b^2-b$

which can be rewritten as: $(a-b)\left(a+b-\frac{1}{2}\right)=0$

So either $a=b$, which gives $A=B$, as $\cos$ is $1:1$ on $\left[0,\frac{\pi}{2}\right]$

Or $a+b=\frac{1}{2}$

But the latter is impossible, because $A+B<\pi$, so at least one of the angles $A$ and $B$ has to be acute, which gives: $a=\cos\frac{A}{2}>\cos\frac{\pi}{4}>\frac{1}{2}$

The whole point is that we look at the interior bisectors only. But if we consider the exterior bisectors as well, it is possible to find $A\ne B$ satisfying $\cos\frac{A}{2}+\cos\frac{B}{2}=\frac{1}{2}$

Answer 2

$$\cos\dfrac A2-\cos\dfrac B2=2\left(\cos^2\dfrac A2-\cos^2\dfrac B2\right)$$

What if $\cos\dfrac A2-\cos\dfrac B2=0?$

else $$\dfrac12=\cos\dfrac A2+\cos\dfrac B2$$

which is clearly possible as $0<\cos\dfrac A2,\cos\dfrac B2<1$